Question

Four grapefruit (considered spheres) 6

 in. in diameter and place in a square box whose inside base dimensions are 12

 in. in the space between the first four grapefruit a fifth of the same diameter is placed. How deep must the box be so that the top will just touch the fifth grapefruit? 

Answer 1

Solution

A,b,c,D,E is the center of the grapefruit ..

So $AB=BC=CD=AD=BE=CE=b$

ABCD is a square

$OC= \frac{AC}2\space BC=3\sqrt{2}$

$h_1= \sqrt{EC^2-OC^2}=\sqrt{36-18}=b\sqrt{2}$

$So \space h=r+h=\frac{1}2 *6+ 3 \sqrt{2}+ \frac{1}2*6$

$=6+3 \sqrt{2} \space in$